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Exercise 4 (Work and Energy) Suggested Solutions 1. (a) (i) Work done by the force = F s = 50 N x 0.1 m = 5 J (ii) Work done by the force = increase in KE 5 = ½ mv2 => 5 = ½ (0.1) v2 => v = 10 m s-1 (b) F = ( mv – mu ) / t => [(0.1)(10) – (0.1)(-10)] / 0.2 = 10 N (c) The first statement is false. By Newton’s first law, a body will move with constant velocity if there is no external force acting on it. Since there is no friction, the disc can therefore move with constant velocity without a driving force. The second statement is true. By Newton’s second law, force is equal to rate of change of momentum. The faster the disc moves toward the wall, the greater the change of momentum. So the average force exerted on the wall would be greater. 2. (c) (i) Acceleration = 8 / 5 = 1.6 m s-2 (ii) Stopping distance = area under graph = ½ x 8 x (15-5) = 40 m (iii) Deceleration = 8 / (15-5) = 0.8 m s-2 Frictional force = ma = 60 (0.8) = 48 N (d) The sledge is released from the same height in both cases. The sledge acquires equal speeds (or KE) at B in both cases. So the stopping distance along BC would remain unchanged. 3. (a) (b) Friction = mg sin = 1 x 10 sin 30o = 5 N (c) a = F / m = 2 / 1 = 2 m s-2 (d) Tilt the runway at a suitable angle such that the trolley will move with a uniform speed after given a sharp push. (e) (i) By W = Fs = 2 1 = 2 J The work done on the trolley by the force is 2 J. (ii) Gain in kinetic energy = work done by the force = 2 J (iii) By KE = 1 2 mv 2 => 2 = 1 1v2 2 => v = 2 m s1 4.(a) Let f be the air resistance. Work done by air resistance = kinetic energy loss of the car 1 f 50 = 500 60 2 2 f = 18 000 N (Alternative method: F = ma, v2 60 2 f=m = 500 = 18 000 N) 2a 2 50 The average air resistance required is 18 000 N. (b) (i) Let h be the height of the plane. By the conservation of energy, KE = PE + work done against resistance 1 Δ mv 2 = mgh + fs 2 1 h 500 (602 202) = 500 10 h + 1500 2 sin 30 h = 100 m (ii) I will use parachute to stop the car. It is because from (i) such inclined plane should be 100 m high and 173 m across, where 100 m is roughly the height of a 28-storey building. It is not practical to use this method. In addition, such inclined plane can only slow down the car, but not stopping the car. 1M 1A 1M 1A 1A 1A 1A MC 1-5 A C E D A 6-10 C A A C B 11-15 B B A A C 16-18 D B A ----------------------------------------------------------------------Explanations to MC 1. Loss in PE = gain in KE mgx = m v2 / 2 x = v2 / 2g = 2.62 / 2 (10) = 0.338 m 2. Work done = increase in KE = (1/2)(5) (102 - 62) = 160 J 4. (1) is wrong because both spheres have the same acceleration. For (2), s = (1/2)gt2 For sphere A, 4h = (1/2)g tA2 For sphere B, h = (1/2) g tB2 So tA = 2tB For (3), when reaching the ground, KE of A = mg (4h) = 1(g) (4h) = 4gh KE of B = mg (h) = 2 g (h) = 2 gh KE of A = 2 KE of B 5. mg (0.3+0.15) = mvB2 / 2 => vB = 3 m s-1 mg (0.3-0.1) = m vA2/ 2 => vA = 2 m s -1 6. Minimum power output = mgh / t = 120 (500) (10) / 60 = 10000 W 7. PE = mgh = mg (H-s) where H is the initial height PE = - mgs + mgH The graph is a straight line with negative slope -mg, and positive intercept mgH. 8. Work done against friction = Fs = F v t = 5 x 10 x 4 = 200 J 9. For (1), PE = mgh, obviously the PE is increasing. For (2), tension = 24 N, mg = 2 x 10 = 20 N, so net force on the block is T - mg = 24 - 20 = 4 N, therefore the block must be accelerating, i.e. the kinetic energy must be increasing. (3) is wrong because average power = (PE gained + KE gained) / time PE gained = 2 x 10 x 4 = 80 J acceleration = F/m = (24 - 20) / 2 = 2 m s-2. speed gained v = u + at = 0 + 2 x 2 = 4 m s-1. So KE gained = mv2/2 = 2 x 42 / 2 = 16 J. => average power = (80+16)/ 2 = 48 W. Alternative faster method : Power developed by the machine = Fs / t = 24 x 4 / 2 = 48 W 10. Useful output power = KE gained / time = [m v2 / 2] / 4 = [m (u+at)2 / 2] /4 = [2000 (3 x 4)2 / 2] /4 = 36 kW 11. Energy loss = KE loss 1 x 0.04 (3302 – 2082) = 1312 J 2 12. Output power = mgh / t = mgv = 1000 x 10 x 0.1 = 1000 W 13. Total energy = PE + KE = PEmax (when KE = 0 at highest point) = mgh = M x 10 x 0.1 = M J. So statement (1) is correct. During swinging down, loss in PE = gain in KE mgh = (1/2) mv2 v 2 gh independent of mass So statement (2) is incorrect. With a pin fixed at C, the bob will rise to the same level as A. So statement (3) is incorrect. 14. Work done against friction = loss in KE 1 m(40) 2 2 1 In case 2, fd 2 m(80) 2 2 In case 1, fd1 So d1 : d2 = 402 : 802 = 1:4 15. v = u + (-10) t = u – 10t KE 1 2 1 2 mv mu 10t (This is a parabola) 2 2 In addition, the KE during the upward journey should decrease to zero at the highest point and then increase as the diver gains speed. 16. Assume initial height = H After traveling for d meters, the skier reaches a height h. Then d sin = H - h Multiply both sides by mg, we get m g d sin = mg (H - h) PE = mgh = mgH – mgd sin PE = - mg sin d + mgH Thus a plot of PE versus d is a straight line with slope – mg sin , and y-intercept mgH. Net work done = work done by net force = increase in KE (mg sin - f) d = KE - 0 KE = (mg sin -f) d So a plot of KE versus d is a straight line with positive slope. Since loss in PE = gain in KE + work done against friction, so the answer is not C, because in option C, the loss in PE is exactly equal to gain in KE, this corresponds to frictionless slope. 17. Engine force = mg sin + friction = mg sin 30o + 0.5 mg = mg Power of car = engine force x velocity = mg v 18. Loss in PE = gain in KE + work done against friction mgh = (1/2) mv2 + fs There is no increase in KE because the block travels with constant speed. So work done against friction = mgh = 1 x 10 x 2 x sin 30o =10 J Supplementary Exercise for Exercise 4 1. The figure below shows a conveying belt system in which a belt is driven by a motor. Cases, each of mass 20 kg, are placed on the belt and delivered up the inclined plane at a uniform speed of 0.5 m s-1. (a) Each case is carried along the inclined plane by the belt without sliding. What is the frictional force acting on each case by the belt ? (b) Find the rate of potential energy gained by each case when it is delivered up to the top of the inclined plane. (c) Suppose the maximum power of the machine is 1000 W, and the efficiency of the conveying belt is 60 %. Calculate the maximum number of cases the belt can carry at the same time at a uniform speed of 0.5 m s-1. Multiple Choice Questions 1. A car stopped after emergency braking. The skid mark left by the car was 22.3 m long. Assume that the friction between the road and tyres was 0.65 times that of the weight of the car. Estimate the speed of the car when it began to skid. A. 5.38 m s-1 B. 12.0 m s-1 C. 16.2 m s-1 D. 17.0 m s-1 2. A man pulls a 50 kg block up a smooth inclined plane making an angle 30o with the horizontal as shown. The inclined plane is 6 m long and the block is pulled from the bottom of the inclined plane to the top in 30 s. Find the average useful output power of the man. A. 5 W B. 10 W C. 50 W D. 87 W 3. Which of the following physical quantities is/are vectors? (1) Force (2) Work done (3) Kinetic energy A (1) only. (2) only. C (1) and (2) only. D (1), (2) and (3). 4. In a marathon, the winner and the runner-up have the same mass. Compared with the runner-up, the winner has more A kinetic energy. B potential energy. Answers for supplementary exercise 7 1. (a) Frictional force = mg sin = 20 (10) (3/6) = 100 N (b) Time required = s/v = 6 / 0.5 = 12 s Rate of PE gained = PE gained / time = 20 (10) (3) / 12 = 50 W (c) Let the required number be n n (50) 1000 (60%) n = 12 MC 1-4 1. fs DCAD 1 2 1 mv 0.65mg 22.3 mv 2 2 2 v = 17.0 m s-1 2. Output power = mgh / t = 50 x 10 x 6 sin 30o / 30 = 50 W C force. D power.